MATH 311-504
Topics in Applied Mathematics
Lecture 2-9:
Basis and dimension (continued).
Matrix of a linear transformation.
Basis and dimension
Definition. Let V be a vector space. A linearly
independent spanning set for V is called a basis.
Theorem Any vector space V has a basis. If V
has a finite basis, then all bases for V are finite and
have the same number of elements.
Definition. The dimension of a vector space V ,
denoted dim V , is the number of elements in any of
its bases.
Examples. • dim Rn = n
• Mm,n (R): the space of m×n matrices
dim Mm,n (R) = mn
• Pn : polynomials of degree at most n
dim Pn = n + 1
• P: the space of all polynomials
dim P = ∞
• {0}: the trivial vector space
dim {0} = 0
Bases for Rn
Let v1 , v2 , . . . , vm be vectors in Rn .
Theorem 1 If m < n then the vectors
v1 , v2 , . . . , vm do not span Rn .
Theorem 2 If m > n then the vectors
v1 , v2 , . . . , vm are linearly dependent.
Theorem 3 If m = n then the following
conditions are equivalent:
(i) {v1 , v2 , . . . , vn } is a basis for Rn ;
(ii) {v1 , v2 , . . . , vn } is a spanning set for Rn ;
(iii) {v1 , v2 , . . . , vn } is a linearly independent set.
Theorem Let S be a subset of a vector space V .
Then the following conditions are equivalent:
(i) S is a linearly independent spanning set for V ,
i.e., a basis;
(ii) S is a minimal spanning set for V ;
(iii) S is a maximal linearly independent subset of V .
“Minimal spanning set” means “remove any element
from this set, and it is no longer a spanning set”.
“Maximal linearly independent subset” means
“add any element of V to this set, and it will
become linearly dependent”.
How to find a basis?
Theorem Let V be a vector space. Then
(i) any spanning set for V can be reduced to a
minimal spanning set;
(ii) any linearly independent subset of V can be
extended to a maximal linearly independent set.
That is, any spanning set contains a basis, while any
linearly independent set is contained in a basis.
Approach 1. Get a spanning set for the vector
space, then reduce this set to a basis.
Approach 2. Build a maximal linearly independent
set adding one vector at a time.
1
1
−1
Example. f : R3 → R2 , f (x) =
x.
2 1 0
Find the dimension of the image of f .
The image of f is spanned by columns of the
matrix: v1 = (1, 2), v2 = (1, 1), and v3 = (−1, 0).
Observe that v3 = v1 − 2v2 . It follows that Im f is
spanned by vectors v1 and v2 alone. Clearly, v1 and
v2 are linearly independent. Hence {v1 , v2 } is a
basis for Im f and dim Im f = 2.
Alternatively, since v1 and v2 are linearly
independent, they constitute a basis for R2 .
It follows that Im f = R2 and dim Im f = 2.
1
1
−1
Example. f : R3 → R2 , f (x) =
x.
2 1 0
Find the dimension of the null-space of f .
The null-space of f is the solution set of the system
1 1 −1
x = 0.
2 1 0
To solve the system, we convert the matrix to
reduced
form:
1 1 −1
1 1 −1
1 0 1
→
→
2 1 0
0 −1 2
0 1 −2
Hence (x, y , z) ∈ Null f if x + z = y − 2z = 0.
General solution: (x, y , z) = (−t, 2t, t), t ∈ R.
Thus Null f is the line t(−1, 2, 1) and dim Null f = 1.
Example. L : P4 → P4 , (Lp)(x) = p(x) + p(−x).
Find the dimensions of Im L and Null L.
p(x) = a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4
=⇒ (Lp)(x) = 2a0 + 2a2 x 2 + 2a4 x 4 .
Since {1, x, x 2 , x 3 , x 4 } is a basis for P4 , the image
of L is spanned by polynomials L1, Lx, Lx 2 , Lx 3 , Lx 4 .
L1 = 2, Lx 2 = 2x 2 , Lx 4 = 2x 4 , Lx = Lx 3 = 0.
Hence Im L is spanned by 1, x 2 , x 4 . Clearly, 1, x 2 , x 4
are linearly independent so that they form a basis
for Im L and dim Im L = 3.
The null-space of L consists of polynomials
a1 x + a3 x 3 . That is, it is spanned by x and x 3 . Thus
{x, x 3 } is a basis for Null L and dim Null L = 2.
Basis and coordinates
If {v1 , v2 , . . . , vn } is a basis for a vector space V ,
then any vector v ∈ V has a unique representation
v = x1 v1 + x2 v2 + · · · + xn vn ,
where xi ∈ R. The coefficients x1 , x2 , . . . , xn are
called the coordinates of v with respect to the
ordered basis v1 , v2 , . . . , vn .
The mapping
vector v 7→ its coordinates (x1 , x2 , . . . , xn )
provides a one-to-one correspondence between V
and Rn . Besides, this mapping is linear.
Matrix of a linear mapping
Let V , W be vector spaces and f : V → W be a linear map.
Let v1 , v2 , . . . , vn be a basis for V and g1 : V → Rn be the
coordinate mapping corresponding to this basis.
Let w1 , w2 , . . . , wm be a basis for W and g2 : W → Rm be
the coordinate mapping corresponding to this basis.
V


g1 y
Rn
f
−→
W


yg2
−→ Rm
The composition g2 ◦f ◦g1−1 is a linear mapping of Rn to Rm .
It is represented as v 7→ Av, where A is an m×n matrix.
A is called the matrix of f with respect to bases v1 , . . . , vn
and w1 , . . . , wm . Columns of A are coordinates of vectors
f (v1 ), . . . , f (vn ) with respect to the basis w1 , . . . , wm .
Examples. • D : P2 → P1 , (Dp)(x) = p ′ (x).
Let AD be the matrix of D with respect to the bases
1, x, x 2 and 1, x. Columns of AD are coordinates
of polynomials D1, Dx, Dx 2 w.r.t. the basis 1, x.
0
1
0
D1 = 0, Dx = 1, Dx 2 = 2x =⇒ AD =
0 0 2
• L : P2 → P2 , (Lp)(x) = p(x + 1).
Let AL be the matrix of L w.r.t. the basis 1, x, x 2 .
L1 = 1, Lx = 1 + x, Lx 2 = (x + 1)2 = 1 + 2x + x 2 .


1 1 1
=⇒ AL = 0 1 2
0 0 1
Problem. Consider a linear operator L : R2 → R2 ,
x
1 1
x
L
=
.
y
0 1
y
Find the matrix of L with respect to the basis
v1 = (3, 1), v2 = (2, 1).
Let N be the desired matrix. Columns of N are coordinates of
the vectors L(v1 ) and L(v2 ) w.r.t. the basis v1 , v2 .
3
2
1 1
4
3
1 1
.
=
, L(v2 ) =
=
L(v1 ) =
1
1
0 1
1
1
0 1
Clearly, L(v2 ) = v1 = 1v1 + 0v2 .
3α + 2β = 4
L(v1 ) = αv1 + βv2 ⇐⇒
α+β =1
2 1
.
Thus N =
−1 0
⇐⇒
α=2
β = −1